Question

Figure shows two processes A and B on a system. Let ∆Q₁ and ∆Q₂ be the heat given to the system in processes A and B respectively. Then ____________ .

विकल्प

• ∆Q₁ > ∆Q₂

• ∆Q₁ = ∆Q₂

• ∆Q₁ < ∆Q₂

• ∆Q₁ ≤ ∆Q₂

Answer 1

∆Q₁ > ∆Q₂

 

Both the processes A and B have common initial and final points. So, change in internal energy, ∆U is same in both the cases. Internal energy is a state function that does not depend on the path followed.

In the P-V diagram, the area under the curve represents the work done on the system, ∆W. Since area under curve A > area under curve B, ∆W_​1> ∆W₂.

Now,

∆Q₁ = ∆U + ∆W₁

∆Q₂ = ∆U + ∆W₂

But ∆W₁ > ∆W₂

⇒ ∆Q₁ > ∆Q₂

Here, ∆Q₁ and ∆Q₂ denote the heat given to the system in processes A and B, respectively.